I am reading the special relativity lecture notes that I got from a professor of mine. It says that the Lagrangian is $$L = \frac{1}{2}m|\dot{\boldsymbol{x}}|^2 - V(\boldsymbol{x}) \tag{1}$$

The notes then go on to say $$\frac{\partial L}{\partial \dot{\boldsymbol{x}}} = m\dot{\boldsymbol{x}} \tag{2}$$

I am familiar with derivatives with respect to a single scalar variable but what does it really mean to have a derivative of a scalar with respect to a vector? Also, how do I reliably get from eqn (1) to eqn (2)?

  • $\begingroup$ Have you considered the three components of vector equation (2)? $\endgroup$
    – Ghoster
    Commented Aug 9, 2023 at 5:01
  • $\begingroup$ $\partial L/\partial x^i=\partial_i f$ is a row vector, or a column vector if our definition of $\partial L/\partial\vec{x}$ implicitly includes a transpose. Similarly for $\partial L/\partial\dot{\vec{x}}$. $\endgroup$
    – J.G.
    Commented Aug 9, 2023 at 7:33
  • 1
    $\begingroup$ Derivative with vectors $\endgroup$
    – Farcher
    Commented Aug 9, 2023 at 7:38
  • $\begingroup$ along certain direction? $\endgroup$
    – freude
    Commented Aug 11, 2023 at 3:42
  • $\begingroup$ en.wikipedia.org/wiki/Directional_derivative $\endgroup$
    – freude
    Commented Aug 11, 2023 at 3:44

1 Answer 1


This notation is really quite confusing, but it comes up because of that pesky dot on top of the vector $\bf{x}$. You get an idea of what this means by interpreting the (slightly simpler) quantity, $$\frac{\partial L}{\partial \bf{x}} = \nabla L$$ In other words, this “vector derivative” is just interpreted as the gradient. In exactly analogous terms, the notation your professor uses is just like taking the gradient, but with respect to the time derivative of the spatial variables rather than the spatial variables themselves. It is a “velocity gradient.” If you follow this through, you will find that because the coordinates and their velocities are independent, the derivative of the potential vanishes, and you are left with, $$\frac{\partial }{\partial {\bf\dot{x}}} V({\bf x}) = 0$$ $$\frac{\partial }{\partial {\bf\dot{x}}} \frac{1}{2} m{\bf \dot{x}}\cdot {\bf\dot{x}} = m \bf{\dot{x}}$$ This can be verified by following through the calculation with components. For example, $$\frac{\partial }{\partial \dot{x}} \frac{1}{2} m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) = m\dot{x}$$ The other components come out likewise, and so the resultant vector really has the components we asserted it did. Unfortunately, I have had to resort to similarly iffy notation at times, as there just isn’t a great standard way to represent that idea. I have seen the notation $\dot{\nabla}$ used before, but it certainly isn’t standard. Hope this helps clarify the notation.


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